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3 years ago

PLEASE HELP! a little confused

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

D and c are parallel
D and a are perpendicular
D and b are perpendicular
C and a are perpendicular
C and b are perpendicular

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What is the value of the missing exponent in the expression below 523 ÷ 10□ = 52.3

Lerok [7]

523% × 10 = (523/100) × 10 = 52.3

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3 years ago

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Two numbers such that the greater number is 75 percent more than the lesser number

victus00 [196]

<h2>Greetings!</h2>

If x = the first number (equal to 1x) then the second number is 75% more than this.

This means the value of the bigger number is 100 + 75 (because 100 is the starting number and 75 is the percent bigger)

This can be shown with the following equation:

Amount * $\frac{percentage}{100}$

x * $\frac{175}{100}$ = 1.75x

1.75x is the value of the bigger number.

<h2>Hope this helps!</h2>

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3 years ago

Use a calculator to approximate the measure of

OLEGan [10]

Answer:

Step-by-step explanation:

I did 90-26 I hope that’s what you’re looking for :))

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3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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3 years ago

Answer a and b ASAP

melomori [17]

Answer:

The area of the rectangle is 48cm2

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